Electronic Devices - DC Biasing-FETs - Discussion
Discussion Forum : DC Biasing-FETs - General Questions (Q.No. 3)
3.
Given the values of VDQ and IDQ for this circuit, determine the required values of RD and RS.
2 k
, 2 k
, 2 k1 k, 5.3 k
, 5.3 k3.2 k, 400
, 400 2.5 k, 5.3 k
, 5.3 kDiscussion:
6 comments Page 1 of 1.
Millize said:
4 years ago
Vdsq = Vd-Vs -> equation 1.
KVL loop at Drain-Source:
20-Idq(Rd+Rs) - Vdsq=0 ; Vdsq = 20-Idq(Rd+Rs). -> equation 2.
Substitute values and equation 2 to equation 1.
20-Idq(Rd+Rs) = 12-IdqRs ; 20-(2.5mA)(3.2k+Rs) = 12-(2.5mA)(Rs).
Solving for Rs;
Rs = 400 ohms.
KVL loop at Drain-Source:
20-Idq(Rd+Rs) - Vdsq=0 ; Vdsq = 20-Idq(Rd+Rs). -> equation 2.
Substitute values and equation 2 to equation 1.
20-Idq(Rd+Rs) = 12-IdqRs ; 20-(2.5mA)(3.2k+Rs) = 12-(2.5mA)(Rs).
Solving for Rs;
Rs = 400 ohms.
Dnzl said:
4 years ago
To solve for Rs, use Shockley's eqn.
Id = IDSS(1-VGS/VP)^2 , Where VGS = -IdRs.
Solving for Rs, Rs=425.4 ohms.
Id = IDSS(1-VGS/VP)^2 , Where VGS = -IdRs.
Solving for Rs, Rs=425.4 ohms.
Mae said:
5 years ago
Then, How to solve the 400 ohms?
(1)
Bitblip said:
6 years ago
20V-2.5mA(Rd)-12Vdq = 0.
Rd = 3200 or 3.2k ohms.
Rd = 3200 or 3.2k ohms.
(1)
ASVITHA said:
7 years ago
v=IR, Then the voltage difference between rd is Vdd-Vdq.
20-12=8v.
then the current IDq=2.5mA.
RD = VDD-VDQ/IDq = 8/2.5m = 3.2*10^3 = 3.2kohm.
20-12=8v.
then the current IDq=2.5mA.
RD = VDD-VDQ/IDq = 8/2.5m = 3.2*10^3 = 3.2kohm.
(1)
SAURABH PANDIT said:
7 years ago
20-12=IDq*RD then calculate RD to get the answer.
(1)
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